Problem:
The vertices of isosceles trapezoid $WXYZ$ have integer coordinates, with $W=(11,12)$ and $Z=(18,36).$ Let the trapezoid have no horizontal or vertical sides, and let $\overline{WX}$ and $\overline{YZ}$ be the only parallel sides. Evaluate the sum of the absolute values of all possible slopes for $\overline{WX}$.
Attempt:
I labeled points $X$ and $Y$ to $(x_{1},y_{1})$, and $(x_{2},y_{2})$ respectively. Then I wrote the equations of lines $\overline{WX}$ and $\overline{YZ}$ to be $$y=mx-11m+12$$ and $$y=mx-18m+36.$$ Therefore, points $X$ and $Y$ have coordinates $(x_{1},mx_{1}-11m+12)$ and $(x_{2},mx_{2}-18m+36)$. Also, since trapezoid $WXYZ$ is isosceles, legs $\overline{XY}$ and $\overline{WZ}$ both equal $$\sqrt{(36-12)^2+(18-11)^2}=25$$ Therefore, $$\sqrt{(x_{2}-x_{1})^2+(mx_{2}-mx_{1}-7m+24)^2}=25.$$ Since $x_1$ and $x_2$ are both integers, we can let $x_2-x_1=c$
$$\implies \sqrt{(c)^2+(mc-7m+24)^2}=25$$ $$\implies c^2+m^2c^2-14m^2c+48mc+49m^2-336m-49=0$$ $$\implies c^2(m^2+1)+c(-14m^2+48)+(49m^2-336m-49)=0$$
At this point, I felt that using the quadratic equation to solve for c would be too bashy. Is there an elegant approach to this problem? Any help is appreciated!
As you did, let $X(x_1,y_1),Y(x_2,y_2)$ and $m$ be the slope of the line $WX$.
You already have $$y_1=mx_1-11m+12\qquad\text{and}\qquad y_2=mx_2-18m+36$$
Now, since we have $$m=\frac{y_1-12}{x_1-11}\quad \text{with}\quad x_1(\not=11),y_1\in\mathbb Z$$ we know that $m$ has to be a rational number.
Let $M,N$ be the midpoint of the side $WX,YZ$ respectively.
Then, $$M\left(\frac{11+x_1}{2},\frac{12+mx_1-11m+12}{2}\right),\quad N\left(\frac{18+x_2}{2},\frac{36+mx_2-18m+36}{2}\right)$$ Since the line $MN$ has to be perpendicular to $WX$, we get $$\dfrac{\dfrac{12+mx_1-11m+12}{2}-\dfrac{36+mx_2-18m+36}{2}}{\dfrac{11+x_1}{2}-\dfrac{18+x_2}{2}}=-\frac 1m,$$ i.e. $$x_2-x_1=7-\frac{48m+14}{m^2+1}\tag1$$ Let $g(m)$ be the RHS of $(1)$. Then, $$g'(m)=\frac{4(4m-3)(3m+4)}{(m^2+1)^2}=0\iff m=\frac 34,-\frac 43$$ from which we see that the range of $g(m)$ is $$-25\le g\left(\frac 34\right)\le g(m)\le g\left(-\frac 43\right)=25$$
$(1)$ is equivalent to $$(c-7)m^2+48m+c+7=0$$ where $c=x_2-x_1\in\mathbb Z$ with $|c|\le 25$ and $c\not=0$.
Now we have $c\not=7$ since for $c=7$, $XY$ is parallel to $WZ$.
With $c\not=7$, we get $$m=\frac{-24\pm\sqrt{25^2-c^2}}{c-7}$$
Since $m\not=0$ has to be a rational number, $\sqrt{25^2-c^2}$ has to be a rational number.
Noticing that $y_1\not=y_2\implies c\not=\pm 25$, we have $$(c,m)=\left(\pm 24,\mp 1\right),\left(24,-\frac{31}{17}\right),\left(-24,\frac{17}{31}\right),\left(20,-3\right),\left(20,-\frac{9}{13}\right),\left(-20,\frac 13\right),$$$$\left(-20,\frac{13}{9}\right),\left(15,-\frac{11}{2}\right),\left(15,-\frac 12\right),\left(-15,2\right),\left(-15,\frac{2}{11}\right),\left(-7,\frac{24}{7}\right)$$
Since each pair is sufficient, the answer is $$1+1+\frac{31}{17}+\frac{17}{31}+3+\frac{9}{13}+\frac 13+\frac{13}{9}+\frac{11}{2}+\frac 12+2+\frac{2}{11}+\frac{24}{7}=\color{red}{\frac{101850442}{4747743}}$$